Physics, Chemistry Mathematics...2 = 23 – 12 = 11 (b) As 23 12 Mgcontains even number of protons (12) against 23 11 Nawhich has odd number protons (11), therefore 23 12 Mghas greater

CodeA

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Physics, Chemistry&

Mathematics

SAMPLE QUESTION PAPER FOR

SOLUTIONS

CBSE Class XII Board Exam 2018

(1)

Physics (Code-A) Solutions of Sample Question Paper for Class XII

SECTION-A

1. radian

2. When one slit is closed, amplitude becomes half and hence intensity becomes one fourth, and there is no

interference.

3. ( ) Y A B A B AB

4. Biconcave lens will change its nature as refractive index of outside medium is greater than that of lens material,

so it will behave as converging lens.

5. (a) Mutual inductance increases on decreasing distance.

(b) Mutual inductance decreases on decreasing the number of turns.

SECTION-B

6. Following are the advantages of digital data over the analog data :

(i) Digital data are immune from imperfection of electronic system.

(ii) By coding the digital data we achieve secure transmission of data.

(iii) Digital data have better noise immunity.

(iv) Digital data can be processed by digital circuit component which are cheap and easily available.

OR

NAND-gate and NOR-gate are the universal gates. They are called universal gates because by using only

NAND-gates or NOR-gates, all the other three gates (OR, AND, NOT-gate) can be formed.

The below combination of NAND-gates works as a OR-gate.

A

B

Y

7. For I = 5 sin(314 t); I0 = 5 A

Iav

= 2 10

5 3.23.14

A

Irms

5 2

0.707 5 3.5352 2

Irms

= 3.535 A

8. KVL based on energy conservation principle while KCL is based upon charge conservation principle.

Physics (Theory) - Class XII (Code-A)

SOLUTIONS

(2)

Solutions of Sample Question Paper for Class XII Physics (Code-A)

9. (a) As, V = – Ir

V

I

As R increases, current (I), in the circuit decreases and terminal voltage (V) increases.

(b) The current across a cell is, I = R r

when R increases, I decreases.

I(A)

R( )

( / ) r

10. Electrons revolve around nucleus and the required centripetal force is provided by attractive force between

electron and nucleus.

2 2

2

kze mv

rr

2

2 kzemv

r

K.E. = 21

2mv =

2

2

kze

r

And potential energy,

U =

2kze

r

Total energy,

E = K.E. + U

E =

2

2

kze

r

SECTION-C

11. p–n junction is developed from a single crystal by introducing donor impurities on one side and acceptor into

the other side. The donor ion is represented by a plus (+) sign because this impurity atom donate an electron

hence it becomes a positive ion. The acceptor ion is indicated by a minus-sign because it accepts an electron

hence it becomes a negative ion.

(3)


Due to density difference across the junction, the holes diffuse toward n-side where the number of holes are

less. Similarly electron diffuse to p-side of the junction. This migration of electron and holes across junction

due to concentration difference is termed as diffusion. At the junction, combination of holes and electrons takes

place since the region of the junction is depleted of the migrating charges it is called the depletion region or

transition region. On the n-side of depletion layer electrons are lost, hence a positive charge build-up on this

side. Similarly the p-side of depletion layer develops a negative charge within depletion layer, there are no mobile

charge carrier. Charge density is positive toward n-side and negative toward p-side. This charge separation

across junction gives rise to an electric field. This electric field is directed from positive charge to negative

charge. Due to this field the positive end of depletion layer is at high potential and the negative end is at low

potential. This potential is called potential barrier or junction barrier which try to oppose any further diffusion

of charge carrier.

If the positive terminal of battery is attached to the p-side and negative terminal of battery toward the n-side,

the field due to this voltage difference compensates the barrier potential and charge carriers are forced to move

across the function. This is called forward bias.

On the other hand, if polarity of battery gets reversed then potential barrier further rises and there is very little

reverse current or practically no flow of current in the circuit.

OR

Vi

Inputvoltage

RB

VBB

I B

IC

VCC

IC

VBE I

E

E

npn

VCE

RC Output

voltage, V0

BC

n-p-n CETransistor in configuration

For Silicon Transistor :

For output circuit V0 = V

CC – I

CRC

…(i)

For input circuit Vi = I

BRB + V

BE…(ii)

(a) As long as, 0.7 Vi

V , there is no collector current.

Then, V0 = V

CC (from equation (i))

The transistor will be in cut-off state.

(b) When 0.7 V 1.0 Vi

V ,

With the increase of Vi beyond 0.7 V, I

C increases almost linearly and V

0 decreases linearly till V

i becomes

1.0 V.

In this situation, the transistor is in active state.

(c) When 1.0 Vi

V ,

The variation of Vi and V

0 is non-linear. In this situation the collector current I

C becomes maximum and the

transistor is in saturation state.

12. Space wave propagation : When the radio waves from the transmitting antenna reach the receiving antenna

either directly or after reflection from the ground or in troposphere, the wave propagation is called space wave

propagation.

(4)


The space wave communication is used in the following communication methods :

(i) For transmitting high frequency TV and FM signals

(ii) Microwaves link and satellite communication.

Let h1 and h

2 be the height of transmitting and receiving antenna respectively, then

1 2h h h

2 1h h h (Range of receiving antenna)

If R is radius of the Earth

12R Rh

trans.

22R Rh

receive =

12 ( )R h h

Total range of transmitting and receiving antenna

1 12 2 ( )R R R Rh R h h

total trans. receive

The range will be maximum; if

1

( ) 0d

Rdh

total

1 1

1

( 2 2 ( )) 0d

Rh R h hdh

⇒

½ ½

1 1

1

( ) 0d

h h hdh

⎡ ⎤⇒ ⎢ ⎥⎣ ⎦

1 1h h h⇒

12

hh⇒

13.

d

ue

A

v0

u0

A

B

A1

B

A

C2

C1

B

Objective

Eye-lens

Magnifying power, m

tan

tanm

2

2

C BA B

C B AB

[A1B = AB]

m = me × m

0

01

e

dm

f

⎛ ⎞ ⎜ ⎟⎝ ⎠

(5)


0

0

1

e

v d

u f

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

0

0

1

e

v dm

fu

⎛ ⎞ ⎜ ⎟

⎝ ⎠

14. For a balanced metre bridge as shown in figure, 100

X l

Y l

, where l = AD = 60 cm

601.5

100 100 60

lX Y Y Y

l

⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦

When 15 resistance is connected in series with Y.

Let Y be the resistance of the series combination of Y and 15 resistance and the balance point be at

a distance of l from end A.

Then Y = Y + 15; l = l – 10 = 60 – 10 = 50 cm

501.5 ( 15) 1.5 15

100 100 50

lX Y Y Y Y Y

l

⎡ ⎤ ⎡ ⎤ ⇒ ⇒ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦

30Y

Also, 1.5 ; 30 1.5 30 45X Y Y X ⇒

when 30 resistance is connected in parallel with Y. Let Y be the resistance of the parallel combination of

Y and 30 resistance then 30 30 900

1530 30 60

Y

If balance point is at a distance of l from end A

100

lX Y

l

⎡ ⎤ ⎢ ⎥⎣ ⎦

45 15100

l

l

⎡ ⎤ ⎢ ⎥⎣ ⎦

3 (100 ) 4 300l l l ⇒ ⇒

75 cml ⇒

15. (a) (i) EM wave propagates in form of varying electric and magnetic fields such that two fields are

perpendicular to each other and also to the direction of propagation of the waves. In other words;

electromagnetic waves are transverse in nature.

(ii) EM waves are produced by accelerated charges.

(iii) In free space, EM wave travels with a velocity, 8 1

0 0

13 10 ms

C .

(iv) EM waves do not require any material medium for their propagation.

(b) (i) Microwaves

(ii) -rays

(iii) X-rays

(6)


16. Since 1

2

2R

R (Where R

1 and R

2 are resistances of upper and lower part)

1 2

2 1

1

2

I R

I R⇒

I2

I1

I I

22 ,I I⇒ and 1 2 1 2

2,

3 3

I II I I I I ⇒

Clearly field due to straight part at centre of circle is zero.

So B due to upper and lower arc

0 2 1

4

I IB

R R

⎡ ⎤ ⎢ ⎥ ⎣ ⎦

0 02 1 1

4 3 3 4 3

⎡ ⎤ ⎢ ⎥⎣ ⎦

I IB

R R, outwards to the plane of paper

17. Let the alternating voltage is E = E0cost.

Let current lags the voltage by angle ,

Where I0 =

0

2

2 1

E

R LC

⎛ ⎞ ⎜ ⎟⎝ ⎠

and

1

tan

LC

R

Average power = 0

T

EIdt

T

∫

= 0 0

0

1( cos ) ( cos( ))

T

E t I t dtT

∫

= 20 0

0

(cos ( )cos cos sin sin ) ∫T

E It t t dt

T

= 0 0

1cos

2E I

average rms rmscosP E I

18. Given, Voltage across RB = 10 V

Resistance RB = 400 k

VBE

= 0, VCE

= 0 RC = 3 k

IB =

Voltage acrossB

B

R

R

=6

3

1025 10 A 25 A

400 10

(7)


Voltage across RC = 10 V

IC = 3

Voltage across 10

3 10

C

C

R

R

= 3.33 × 10–3 A = 3.33 mA

3

6

3.33 10

25 10

C

B

I

I

= 1.33 × 102 133

19. Given, for the first condition, wavelength of light = 600 nm

and for the second condition, wavelength of light = 400 nm

Also, maximum kinetic energy for the second condition is equal to the twice of the kinetic energy in first

condition.

i.e.,max max

2K K

Here,max

hc

K

max 0

2hc

K

1240 12402 1240 eV nm

600 400

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∵ �hc

1240

1.03 eV1200

20. In case of AM, the instantaneous voltage of carrier waves is varied by the modulating wave voltage. So, during

the transmission, noise signals can also be added and receiver assumes noise a part of the modulating signal.

In case of FM, the frequency of carrier waves is changed as the change in the instantaneous voltage of

modulating waves. This can be done by mixing at transmitting end and not while the signal is propagating

through the channel. So noise does not effect FM signal.

21. (a) According to question, a nuclide 1 is said to be mirror isobar of nuclide 2, if Z1 = N

2, Z

2 = N

1.

Now in 11

Na23, Z1 = 11 and N

1 = 23 – 11 = 12

Mirror isobar of 11

Na23 is 12

Mg23, for which Z2 = 12 = N

1 and N

2 = 23 – 12 = 11

(b) As 23

12Mg contains even number of protons (12) against

23

11Nawhich has odd number protons (11), therefore

23

12Mg has greater binding energy than

11Na23.

22. An inductor opposes flow of current through it by developing a back emf according to Lenz’s law. The induced

voltage has a polarity so as to maintain the current at its present value. If the current is decreasing, the polarity

of the induced emf will be so as to increase the current and vice-versa.

(8)


Since, the induced emf is proportional to the rate of change of current, it will provide greater reactance to the

flow of current if the rate of change is faster, i.e., if the frequency is higher. The reactance of an inductor,

therefore, is proportional to the frequency. Mathematically, the reactance offered by the inductor is given by

XL = L.

SECTION-D

23. (a) 50 Hz, 220 V RMS.

(b) No

(c) Mutual Induction

SECTION-E

24. (a) 3 F

3 F 1 F 2 F

2 F

100 V

⇓ equivalent

2 F

3 F

3 F 2 F

1 F

100 V

C1 = 3 F + 3 F = 6 F, C

2 = 1 F + 2 F = 3 F

1 2

12

1 2

6 32

6 3

C CC

C C

F

Finally equivalent capacitor C12

= 2 F and 2 F are in parallel.

Net capacitance C1 = 2 F + 2 F = 4 F

q = CV = 4 × 10–6 × 100 = 4 × 10–4 C

Also, total energy stored

(9)


2 6 21 14 10 100 0.02 J

2 2U CV

q = charge = 4 × 10–4 C

U = 0.02 J

(b) The plate B is earthed

V = potential difference = E d =

0

d⎛ ⎞ ⎜ ⎟⎝ ⎠

Q –Q

BA

0

QV d

A

0AQ

CV d

⇒

OR

(a)

+ + + + + + + + + + + + + + + + + + + + + + +

E

l

Figure shows a small part of infinitely long cylindrical non conducting wire.

Using

0

.

inq

E dA ∫

�

�

0

0 cos0I

EA

0

(2 )l

E rl

⇒

02

Er

⇒

(b) Since electric flux has only x-component. (electric flux) through four faces (other than M and N)

= ExS cos /2 = 0

Ex = 5Ax + 2B

Magnitude of electric field at face M (x = 0)

EM

= 5 × 10 × 0 + 2 × 5 = 10 NC–1

at x = 0.1 m

EN = 5 × 10 × 0.1 + 2 × 5 = 15 NC–1

(10)


M through face M

M

= EM

.A cos = 10 × (0.1 × 0.1) × cos180°

= 0.10 NmC–1

N

= EN.A cos = 15 × 0.1 × 0.1 × cos0°

= 0.15 NmC–1

Net flux through cube

= N –

M = 0.15 – 0.10 = 0.05 NmC–1

From Gauss’s theorem,

0 q = 8.85 × 10–12 × 0.05

q = 0.443 × 10–12 C

25. The path difference between two rays coming from holes S1 and S

2 is (S

2P – S

1P).

X

x –

x +

d

2d

2

P

O

D

d

S1

S2

Now, (S2P)2 – (S

1P)2 =

2 2

2 2

2 2

d dD x D x

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦= 2xd, where S

1S

2 = d and OP = x

(S2P + S

1P) (S

2P – S

1P) = 2xd

2 1

2 1

2xdS P S P

S P S P

If x, d << D, then negligible error will be introduced if (S2P + S

1P) in the denominator is replaced by 2D.

S2P – S

1P =

xd

D

For maximum, S2P – S

1P = n

Thus, xd

nD

Or, ,

n

n Dx x

d

n = 0, ±1, ±2, ±3, ... [For maxima]

Now, for minimum, S2P – S

1P = (2n – 1)

2

(11)


Thus (2 1)2

xdn

D

Or, (2 1)2

n

Dx x n

d, n = ±1, ±2, ±3, ... [For minima]

Thus, bright and dark bands appear on the screen. Such bands are called ‘fringes’. These dark and bright

fringes are equally (spaced).

Expression for fringe width ()

Let nth order bright fringe is at a distance xn an (n + 1)th order bright fringe is at x

n+1 from O. Then

1

( 1)and

n n

n D n Dx x

d d

Now the fringe width is

1n n

Dx x

d

Thus, the expression for fringe width is D

d

OR

(a) Let magnetic field due to current element Idy at P is given by

0

3

( ) sin(90 )

4

I dy rdB

r

0 0

2 2 2

cos cos

4 4 ( )

I dy I dy

r y d

We have,

tany

d

2secdy d d⇒

90° +

p

d

yr

I

dy

2

0

2 2

cos sec

4 sec

I d ddB

d

⇒

0 (cos )4

Id

d

0cos

4

IB dB d

d

⇒

∫ ∫

0 (sin sin )

4

I

d

(b) For infinite wire, 2

0sin sin

4 2 2

IB

d

⎛ ⎞⇒ ⎜ ⎟⎝ ⎠

0 2

4

I

d

(12)


For semi-infinite wire, 0,2

0sin0 sin

4 2

IB

d

⎛ ⎞ ⎜ ⎟⎝ ⎠

0

4

IB

d

⇒

26. (a) Device ‘X’ is a capacitor.

(b) Curve A shows power consumption over a full cycle, curve B shows voltage and curve C shows current.

As in a perfect capacitor the current leads the voltage by a plane angle of /2.

(c) Z = XC =

1

C

1

2C

C

Xf

XC

f1

CX

f

(d) Consider a capacitor with capacitance C be connected to an AC source with an emf having instantaneous

value.

C

E

E = E0sint ... (i)

Due to this emf, charge will be produced and it will charge the plates of capacitor with positive and negative

charge. If potential difference across the plates of capacitor is V, then

orq

V q CVC

The instantaneous value of current in the circuit

( )( )

dq d CEI V E

dt dt

0 0

( sin ) ( sin )d

CE t E E tdt

= 0cosCE t

= 0

cos [ cos sin( / 2 )]1/

C

Et t t

I = 0

sin( / 2)1/

C

Et

... (ii)

(13)


I will be maximum when sin(t + /2) = 1 so that I = I0

Where, peak value of current I0 =

0

1/C

E

I = I0 sin (t + /2) ... (iii)

From (i) and (iii) it is clear current leads the voltage by a phase angle of /2.

t

E Iand

E

I

ORAC generator

AxleCoil

SN

Sliprings Alternating emf

Carbonbrushes

Let, = t is the angle made by area vector of coil with magnetic field.

Magnetic flux linked with coil is given by,

= NBAcos = NBAcost

sind

NBA tdt

sind

NBA tdt

By Faraday’s law of EMI, e = d

dt

Induced emf is given by,

e= NBAsint

e= e0sint

Where, e0= NBA= peak value of induced emf

(14)


Induced

emf

0

O T/4 T/2 3 /4T

90° 180° 270° 360°Time

The mechanical energy spent in rotating the coil in magnetic field appears in electrical energy.

� � �

(15)

Chemistry (Code-A) Solutions of Sample Question Paper for Class XII

SECTION-A

1. When constituent particles are present only at the corner positions of an unit cell, it is called primitive unit

cell.

2. Foam

For example - Whipped cream

3. 2-Phenylethan-1-ol

4. K3[Al(ox)

3] or K

3[Al(C

2O

4)3]

5. CH — C — Br3

CH3

CH3

would give faster SN1 reaction than CH

3 – CH

2 – Br.

SECTION-B

6. Q = It = 0.5 × 2 × 60 × 60 = 3600 C

Electrons flowing through the wire on passing charge 1 F (96500 C) = 6.02 × 1023

Electrons flowing through the wire on passing 3600 C charge =

23226.02 10

3600 2.24 1096500

7. (i) C6H

5CH

2NH

2 > C

6H

5NHCH

3 > C

6H

5NH

2

(ii)

NH2

CH3

>

NH2

>

NH2

NO2

8. (i) COCl 6NH + AgNO3

3 3AgCl

(1 mol) (3 mol)

Structural formula of complex = [Co(NH3)6]Cl

3

(ii) Hexamminecobalt (III) chloride.

9. (i) Order of reaction = zero

(ii) Unit of k = mol L–1 sec–1

10. (i)

S OHO

OHTriangular pyramidal

(ii) Xe

O

Square pyramidalF

F

F

F

Chemistry (Theory) - Class XII (Code-A)

SOLUTIONS

(16)

Solutions of Sample Question Paper for Class XII Chemistry (Code-A)

OR

(i)

P

OHHO

OH

Tetrahedral

O

(ii)Xe

O

Triangular pyramidalOO

SECTION-C

11. d = 3

A

ZM

a N

7.5 = 10 3 23

2 M

(500 10 ) 6.02 10

M = 282 g/mol

Number of moles = 300

282

Number of atoms = 23300

6.02 10282

= 6.4 × 1023

12. (i) Red phosphorus is less reactive than white phosphorus as white phosphorus has greater angle strain in

P4 molecule.

(ii) Because halogens have high effective nuclear charge and smallest size in respective period.

(iii) In N2O

5, the oxidation state of N is +5 whereas in N

2O

3, it is +3.

Higher the oxidation state of central atom in nitrogen oxide, higher will be acidic character.

13.

2

s urea

H O

P° – P n

P° n

s23.8 P

23.8

=

2

2

urea H O

urea H O

W MW

MW W

s23.8 P

23.8

=

30 18

60 846

23.8 – Ps = 0.0106 × 23.8 = 0·253

Ps = 23.8 – 0.253 = 23.547 mm Hg

14. Ideal solution

(i) Obeys Raoult's law at every range of concentration.

(ii) Neither the heat is absorbed nor evolved.

Non-ideal solution

(i) They do not obey Raoult's law.

(ii) Heat is evolved or absorbed during dissolution.

(17)


15. A0 = 100%

A = (100 – 25)% = 75%

t = 20 min.

k = 0A1

2.303 logt A

k = 1 100

2.303 log20 75

= 0.0144 min–1

For 75% completion of reaction,

t = 0A1

2.303 logAk

t = 1 100

2.303 log0.0144 25

= 96.3 min

16. (a) These are sodium salt of sulphonated long chain alcohols or hydrocarbon e.g., :- Sodium dodecylbenzene

sulfonate.

(b) The antibiotics which kill or inhibit a short range of gram-positive or gram-negative bacteria are known as

“narrow spectrum antibiotics” but if they are effective against a single organism or disease, they are

referred to as “limited spectrum antibiotics”.

17. (a) Nylon-6, 6

NH2 – (CH

2)6 – NH

2COOH – (CH

2)4 – COOH

Hexamethylene diamine Adipic acid

(b) Melamine-formaldehyde polymer

NH N

2

NH2

NH2

N

Melamine

N

HCHO

Formaldehyde

(c) Buna-S

2 2

1.3 butadiene

CH CH CH CH

CH CH2

Styrene

OR

(a) Buna-N

CH = CH – CH = CH + CH = CH2 22

CH1, 3-Butadiene Acrylonitrile

(b) Natural rubber

CH = C – CH = CH2 2

CH3

2-methyl-1, 3-butadiene

(c) Teflon

CF2 = CF

2

Tetrafluoroethylene

(18)


18. (a) CH3COOH

NH /3

CHCl alc.KOHBr /KOH 323 2 3 2 3

A B C

CH CONH CH NH CH NC

(b) C6H

5N

2

+BF4

– NaNO /Cu2

NO2

(A)

Fe/HCl

NH2

(B)

CH COCl/Pyridine3

NHCOCH3

(C)

19. (i) CH – CH – C – CH3 2 3

Br

CH3

(ii) CH3 – CH

2 – CH = CH – CH

3

(iii)

CH3

CH3

CH3

CH3

+

20. (i) van-Arkel method.

Ni (impure) + 4CO

4

Volatile

Ni(CO) Ni (pure) + 4CO

(ii) (a) To decrease the melting point of mixture.

(b) To increase electrical conductivity of Al2O3.

(iii) Limestone (CaCO3) acts as flux in the extraction of iron from its oxides.

CaCO3

CaO + CO2

2 3(Impurity) (Slag)

CaO + SiO CaSiO

21. (i) CH – C – I + CH OH3 3

CH3

CH3

(ii) CH – CH – C – CH3 2 3

O

(iii)

CHO

OH

22. (i) (CHOH)4

(CHOH)4

CHO CH = N – OH

CH OH2

CH OH2

NH OH2

(ii) In neutral amino acid, –COOH loses a proton and –NH2 of same molecule accept a proton. Due to this

reason, amino acids are amphoteric in nature.

(iii) Because vitamin C is water soluble and it is readily excreted in urine.

(19)


23. (i) Caring and true friendship.

(ii) Drugs that are used to neutralize acidity in stomach are called antacids. For example, NaHCO3 and

Mg(OH)2.

(iii) No, because excessive HCO–

3 can make the stomach alkaline and trigger the production of even

more acid.

SECTION-D

24. (i) (a) Due to lanthanoids contraction.

(b) In case of transition element ns and (n – 1)d electron both participate in bonding due to less

energy difference when ns electron take place in bonding they exhibit lower oxidation state while

in case of higher o.s. both orbital electrons involve in bonding.

(ii) (a) 2 KMnO4 K2MnO4 + MnO2 + O2

(b) 2 Na2CrO4 + 2H+ Na2Cr2O7 + 2Na

+ + H2O

OR

(i) (a) Stability of ions in aqueous state depends on electrode potential because due to small size Cr+3

is more stable.

(b) Mn+3

is strong oxidising agent as Mn+2

is more stable than Mn+3

due to its half filled

configuration.

(c) Ti+4

is colourless ion due to d0 configuration of ion.

(ii) Similarity :

* Both series element exhibit mainly +3 oxidation state.

Difference :

Lanthanoids have less tendency of complex formation whereas actinoids have high tendency of

complex formation.

25. E°cell = 0.34 + 2.37 = 2.71

2

cell cell2

Mg0.06E E log

n Cu

⎡ ⎤⎣ ⎦

= 0.06

2.71 log102

= 2.68 V

G = –nFEcell = –2 × 96500 × 2.68 = –5.17 × 105 J

OR

(XCl) = (X+) + (Cl

–)

= 73.5 + 76.5

= 150

2

1000 1000 2.48 10XCl 124

M 0.2

0

XCl 1240.83

150XCl

(20)


26. (a) (i) A is CH3CHO

B is CH3CH = N – OH

(ii) A is CH3COOH

B is CH3COCl

(b) (i) Iodoform test

(ii) Tollen’s test

(c) CH – C – CH < CH CH OH < CH COOH3 3 3 2 3

O

OR

(a)

CH3

CHO

CrO Cl2 2

(b) C6H

5COCH

3 < CH

3CHO < HCHO

(c) Due to –I effect of Cl

(d) CH3CH

2CH = CH – CH

2CHO

(e) C H O3 6

I + NaOH2

NaOH + I2

CH CCH3 3

CH CH CHO3 2

O(A)

(B)

CHI3

No ppt.

Yellow

� � �

(21)

Mathematics (Code-A) Solutions of Sample Question Paper for Class XII

SECTION-A

1. a b�

�

= ˆ ˆ ˆ ˆ ˆ ˆ4 2 2i j k i j k

= ˆ ˆ ˆ6 3 2i j k

Unit vector parallel to a b�

�

=

ˆ ˆ ˆ6 3 2

36 9 4

i j k

= ˆ ˆ ˆ6 3 2

7

i j k

2. 1tan 3

= 3

1cot 3

= 6

Hence, 3 6 2

⎛ ⎞ ⎜ ⎟⎝ ⎠

3. A2 = A

7A – (I + A)3

7A –[(I + A)2(I + A)] = 7A – [(II + AA + 2AI) (I + A)]

= 7A – [I + A2 + 2AI] [I + A]

= 7A – [I + A + 2A] [I + A]

= 7A – [I + 3A] [I + A]

= 7A – [I I + IA + 3AI + 3A2]

= 7A – [I + A + 3A + 3A]

= – I

4.

2 2sin cos

sin cos

x xdx

x x

∫

= cos2

2sin2

xdx

x

∫

= 2 cot2xdx ∫

= 2log | sin2 |

2

x+ C

= – log | sin 2x | + C

SECTION-B

5.3 7 8 7

2 4 6 4

x

On expanding both determinants we get

MathematicsMathematicsMathematicsMathematicsMathematics - Class XII (Code-A)

SOLUTIONS

(22)

Solutions of Sample Question Paper for Class XII Mathematics (Code-A)

12x + 14 = 32 – 42

12x + 14 = –10

12x = –24

x = –2

6. R(x) = 3x2 + 36x + 5

MR = dR

dx= 6x + 36

When x = 5

MR = 30 + 36 = 66

7. Let c�

be any vector perpendicular to anda b

��

then ( )c a b �

� �

ˆ ˆ ˆ ˆ ˆ ˆ{( 2 3 ) (3 2 )}c i j k i j k �

1 2 3 11 7

3 1 2

i j k

c i j k

�

Now vector of magnitude 171 in the direction of c�

is given by

= magnitude ˆc

= ( 11 7 )

1711 121 49

i j k

= i – 11j – 7k

8. f(x) =

0

sin

x

t tdt∫

f (x) = 1 x sin x – 0

= x sin x

9. y = x

3

dy

dx

⎛ ⎞⎜ ⎟⎝ ⎠ +

2

2

d y

dx

Order of equation = 2

Degree of equation = 1

Hence sum of order and degree = 3

10. Given A2 = kA

1 1 1 1 1 1

1 1 1 1 1 1k

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

2 22

2 2

k kk

k k

⎡ ⎤ ⎡ ⎤ ⇒ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦

11.2 2

1 1x y dx y x dy = 0

2 2

1 1

x ydx dy

x y

= 0

Integrating we get, 2 2

1 1

x ydx dy

x y

∫ ∫ = 0

2 21 1x y = c

(23)


12. I =

4

2

21

xdx

x ∫

Put x2 + 1 = t

2xdx = dt at x = 2

xdx = 1

2dt t = 5

at x = 4

t = 17

I =

17

5

1/ 2

∫ dtt

= 175

1 1 17log | | log

2 2 5t

SECTION-C

13. Prob. of step forward p = .4

Prob. of step backward q = .6

So by binomial distribution

(p + q)5 = 5C0.p5 + 5C

1.p4.q1 + 5C

2.p3.q2 + 5C

3.p2.q3 + 5C

4.p1.q4 + 5C

5.q5

Prob. for man so that he is one-step away from starting point will be

= 5C2.p3.q2 + 5C

3.p2.q3

= 10.(.4)3.(.6)2 + 10.(.4)2.(.6)3

= 10(.16).(.36)

= (1.6).(.36)

= .576

OR

E1

E2

E1 is the event that on throwing a dice 1 or 2 comes

So 1

2 1( )

6 3P E

E2 is the event that on throwing a dice 3, 4, 5, 6 comes

So 2

4 2( )

6 3P E

If A is event that on throwing coin only one tail is obtained.

Then for Prob. of exactly one tail if he got 3, 4, 5 or 6 is

2E

PA

⎛ ⎞⎜ ⎟⎝ ⎠ =

2

2

1 2

1 2

( )

( ) ( )

AP E P

E

A AP E P P E P

E E

⎛ ⎞ ⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

AP

E

⎛ ⎞⎜ ⎟⎝ ⎠ =

1,

2

1

AP

E

⎛ ⎞⎜ ⎟⎝ ⎠

= 3

8

(24)


2

AP

E

⎛ ⎞⎜ ⎟⎝ ⎠ =

2 1

3 2

2 1 1 3

3 2 3 8

=

1

3

1 1

3 8

= 1

3×24

11=

8

11

14. We have,

P.V. of A = ˆ ˆ ˆ3 6 9i j k

P.V. of B = ˆ ˆ ˆ2 3i j k

P.V. of C = ˆ ˆ ˆ2 3i j k

P.V. of D = ˆ ˆ ˆ4 6i j k

AB

��

= ˆ ˆ ˆ2 4 6i j k

AC

��

= ˆ ˆ ˆ3 8i j k

AD

��

= ˆ ˆ( 9)i k

Now, AB AC AD ��

2 4 6

1 3 8 0

1 0 ( 9)

–2(–3 + 27) + 4(– + 9 + 8) –6(0 + 3) = 0

6 – 54 – 4 + 68 – 18 = 0

2 – 4 = 0

= 2

, ,AB AC AD

��

are coplanar and so the points A, B, C and D are coplanar.

15. If the given lines are intersecting then the shortest distance between the lines is zero and also they have same

common point ˆ ˆ ˆ ˆ ˆ ˆ3 2 4 ( 2 2 )r i j k i j k �

3

1

x =

2

2

y =

4

2

z =(Let)

Let P is ( + 3, 2 + 2, 2 – 4)

Also, ˆ ˆ ˆ ˆ ˆ5 2 (3 2 6 )r i j i j k �

5

3

x =

2

2

y =

0

6

z =(Let)

Let Q is (3 + 5, 2 – 2, 6)

If lines are intersecting then coordinate of point P and Q will be same

+ 3 = 3 + 5 ..... (1)

2 + 2 = 2 – 2 ..... (2)

2 – 4 = 6 ..... (3)

Solve (2) and (3)

+ 1 = – 1

– 2 = 3 – + –

3 = –2 – 1

4 = –2

2

Put = –2 ..... (3)

2 – 4 = 6(–2)

(25)


2 = –12 + 4

2 = –8

4 Put and in (1)

+ 3 = 3 + 5

–4 + 3 = 3(–2) + 5

–1 = –1

from = –4 then P is (–1, –6, –12)

from = –2 then Q is (–1, –6, –12)

as P and Q are same

Lines are intersecting lines and their point of intersection is (–1, –6, –12)

OR

A(2, 1, –1) ; B(–1, 3, 4)

90°

A B

AB

��

= OB OA��

AB

��

= ˆ ˆ ˆ3 2 5i j k Given plane x – 2y + 4z = 10

1

n

�

= ˆ ˆ ˆ2 4i j k

The required plane is perpendicular to given plane.

n

�

of required plane will be perpendicular to 1

n

�

and AB

��

1

n n AB��

�

1n

�

= i – 2j + 4k

AB

��

= ˆ ˆ ˆ3 2 5i j k

1n AB

��

�

= ˆ ˆ ˆ18 17 4i j k

Required plane is

r n� �

= a n� �

ˆ ˆ ˆ( 18 17 4 )r i j k �

= ˆ ˆ ˆ ˆ ˆ ˆ(2 ) ( 18 17 4 )i j k i j k

ˆ ˆ ˆ( 18 17 4 )r i j k �

= –36 – 17 + 4

ˆ ˆ ˆ18 17 4 49r i j k �

18 17 4 49x y z

16. I = 5( 3)

dx

x x ∫

I =

4

5 5( 3)

x dx

x x ∫

(26)


Let x5 = t 5x4dx = dt

I = 1

5 ( 3)

dt

t t ∫

I = 1 1 1 1

5 3 3dt

t t

⎛ ⎞ ⎜ ⎟⎝ ⎠∫

I = 1log log( 3)

15t t c

I = 1log

15 3

tc

t

⎛ ⎞ ⎜ ⎟⎝ ⎠

I =

5

5

1log

15 3

xc

x

⎛ ⎞⎜ ⎟⎝ ⎠

17. Let 11 3

sin2 4

= then

3

4= sin 2

Now, tan11 3

sin2 4

⎛ ⎞⎜ ⎟⎝ ⎠ = tan

If sin 2 = 3

4 then

2

2tan

1 tan

= 3

4

8 tan = 3 + 3tan23 tan2 – 8 tan + 3 = 0

tan = 8 64 4 3 3

6

= 4 7

3

tan11 3

sin2 4

⎛ ⎞⎜ ⎟⎝ ⎠ =

4 7

3

Hence proved.

18.dy

dx= (1 + x) + y(1 + x)

or,dy

dx= (1 + y)(1 + x)

or,1

dy

y = (1 + x) dx

1

dy

y∫ = (1 )x dx∫

log |1 + y| = x +

2

2

x

+ C

Given, y = 0 when x = 1

i.e., log |1 + 0| = 3

2 + c

C = 3

2

(27)


The particular solution is

log |1 + y | =

2

2

x

+ x–3

2

or, the answer can expressed as

log |1 + y | =

22 3

2

x x

or, 2

2 3 /21

x x

y e

or, 2

2 3 /21

x x

y e

19. I =

22

2

0

cos

1 3sin

xdxdx

x

∫

I =

22

2

0

1 sin

1 3sin

xdx

x

∫

I =

2

2

0

1 4 1

3 3 (1 3sin )dx

x

⎡ ⎤ ⎢ ⎥⎣ ⎦∫

I =

2 2

2

0 0

1 4 1

3 3 1 3sindx dx

x

∫ ∫

I =

22

2

0

1 4 sec

3 2 3 1 4tan

xdx

x

∫

Let 2 tan x = t

sec2x dx = 1

2dt

I = 2

0

4 1

6 3 2 1

dt

t

∫

I = 1

0

2tan

6 3t

I = 2

6 3 2

I = 6 3

I = 6

20. y = sin–1

12 3

1 (36)

x x

x

⎡ ⎤⎢ ⎥⎣ ⎦

y = sin–12 2 3

1 (36)

x x

x

⎡ ⎤ ⎢ ⎥⎣ ⎦

y = sin–1 2

2 (6)

1 (6)

x

x

⎡ ⎤⎢ ⎥⎣ ⎦

(28)


y = 2 tan–1(6)x

dy

dx= 2

26 log6

1 (6)

x

x

dy

dx =

2 6 log6

1 (36)

x

x

21.

1 1, if 1 0

( )2 1

, if 0 11

kx kxx

xf x

xx

x

⎧ ⎪⎪ ⎨⎪

⎪⎩ function f(x) is continuous at x = 0

f(0) = 0

lim ( )x

f x

0 1

0 1

= 0

1 1limx

kx kx

x

⎛ ⎞ ⎜ ⎟⎝ ⎠

–1 = 0

1 1 1 1lim

1 1x

kx kx kx kx

x kx kx

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

–1 =

0

1 1lim

1 1x

kx kx

x kx kx

–1 = 0

2lim

1 1x

k

kx kx

–1 = 2

2

k k = –1

OR

x = a cos3 and y = a sin3

dx

d = –3a cos2sin and

dy

d = 3a sin2 cos

dy

dx =

dy

d

dx

d

dy

dx = –tan

2

2

d y

dx= –sec2

d

dx

2

2

d y

dx =

2

2

1sec

3 cos sina

2

2

d y

dx =

41sec cosec

3a

2

2

6

d y

dx

⎛ ⎞⎜ ⎟⎝ ⎠

=

4

1 22

3 3a

⎛ ⎞ ⎜ ⎟⎝ ⎠=

32

27a

(29)


22. We have,

dy

dx=

1

2

tan

1

x y

x

21

dy y

dx x

=

1

2

tan

1

x

x

I.F = 2

1

1dx

xe

∫ = 1

tan x

e

1tan

.

x

y e

= 1

1

tan

2

tan

1

xx

e dxx

∫

Put t = tan–1x

dt = 2

1

1

dx

x

=

I I I

tt e dt∫

= 1t t

t e e dt ∫1

tan x

y e

= t et – et + c

1tan x

y e

= (tan–1 x – 1) 1

tan x

e c

y = tan–1 x – 1 + 1

tan x

c e

23. f : R+ [4, )

f(x) = x2 + 4

f(x) = 2x > 0 (one - one)

As, f(x) = x2 + 4 4

Range = [4, ) = co-domain

onto

So f is invertible.

Further : y = x2 + 4

y – 4 = x2 x = ± 4y

As x > 0 so x = 4y

y = 4x = f –1(x)

Or, f–1(y) = 4y

SECTION-D

24. If zmax

= 100x + 120y

Type A Type B

Worker

Capital

2

3

3

1

30

17

Subject to,

(30)


2x + 3y 30

3x + y 17

x 0

y 0

Let object of type A = x

Object of type B = y

y

D(0, 17)

(0, 10) C E (3, 8)

A(0, 0) (15, 0)

B

(2 + 3 30)x y (3 + ) 17x y

n

Coordinate

O

E

C

A

(0, 0)

(3, 8)

(0, 10)

pts. z = 100 + 120max

x y

z = 0

z = 300 + 960 = 1260

z = 1200

173

,0 z =1700

3

maximum revenue = 1260

25.

2

2

2

( )

( )

( )

x y zx zy

zx z y xy

zy xy z x

R1 zR

1, R

2 xR

2, R

3 yR

3

We get

2 2 2

2 2 2

2 2 2

( ) 1

( )

( )

z x y z x z y

zx x z y x y

xyz

zy xy y z x

Taking z, x, y common from C1, C

2, C

3

respectively, we get

2 2 2

2 2 2

2 2 2

( )

( )

( )

x y z z

xyzx z y x

xyz

y y z x

C1 C

1 – C

3, C

2 C

2 – C

3

(31)


2 2 2

2 2 2

2 2 2 2 2

( ) 0

0 ( )

( ) ( ) ( )

x y z z

z y x x

y z x y z x z x

2

2

2

( )( ) 0

0 ( )( )

( )( ) ( )( ) ( )

x y z x y z z

z y x z y x x

y z x y z x y z x y z x z x

2

2 2

2

0

( ) 0

( )

x y z z

x y z z y x x

y z x y z x z x

R

3 R

3 – R

2 – R

1

2

2 2

0

0

2 2 2

x y z z

x y z z y x x

x z zx

1 1 3 2 2 3

1 1,C C C C C C

z x

2

2

22 2

0 0 2

zx y z

x

xx y z z y x

z

zx

Expanding along R3

2xz(x + y + z)2 2 2

x zx y z y

z x

⎛ ⎞ ⎜ ⎟⎝ ⎠

2xz(x + y + z)2 (xz + xy + yz + y2 – xz)

2xyz(x + y + z)3 = RHS

26. P1 is ˆ ˆ3 6 0r i j

�

P1 is x + 3y – 6 = 0

P2 is ˆ ˆ ˆ3 4r i j k

�

P2 is 3x – y – 4z = 0

Equation of plane passing through intersection of P1 and P

2 is P

1 + P

2 = 0

(x + 3y – 6) + (3x – y – 4z) = 0

(1 + 3)x + (3 – )y + (–4)z + (–6) = 0

Its distance from (0, 0, 0) is 1.

2 2 2

0 0 0 61

1 3 3 4

36 = (1 + 3)2 + (3 – )2 + (–4)2

36 = 1 + 92 + 6 + 9 + 2 – 6 + 162

36 = 262 + 10 262 = 26 = ± 1

Hence required plane is

For = 1, (x + 3y – 6) + 1(3x – y – 4z) = 0

(32)


4x + 2y – 4z – 6 = 0

For = –1, (x + 3y – 6) – 1(3x – y – 4z) = 0

–2x + 4y + 4z – 6 = 0

OR

P1 is ˆ ˆ ˆ2r i j k

�

= 5

1

n

�

= ˆ ˆ ˆ2i j k

P2 is ˆ ˆ ˆ3r i j k

�

= 6

2n

�

= ˆ ˆ ˆ3i j k

The line parallel to plane P1 and P

2 will be perpendicular to

1 2and .n n

� �

1 2||b n n

��

1n

�

= ˆ ˆ ˆ2i j k

2n

�

= ˆ ˆ ˆ3i j k

1 2n n� �

= ˆ ˆ ˆ3 5 4i j k

b�

= ˆ ˆ ˆ3 5 4i j k Point is (1, 2, 3).

a

�

= ˆ ˆ ˆ2 3i j k

required line is r a b �

� �

ˆ ˆ ˆ ˆ ˆ ˆ2 3 3 5 4r i j k i j k �

27. 5 Red2 Black

(i) = 0, 1, 2(ii) yes, is a random variable it varies from 0 to 2

x

x

urn.

No. of blackballs = x

P x( )

0 1 2

2042 20

42 2

42

(a) P(0) = P(R1) × P

2

1

R

R

⎛ ⎞⎜ ⎟⎝ ⎠ =

5 4

7 6 =

20

42

(b) P(1) = P(R1) × P

2 2

1

1 1

( )B R

P B PR B

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 5 2 2 5

7 6 7 6 =

20

42

(c) P(2) = P(B1) ×

2

1

BP

B

⎛ ⎞⎜ ⎟⎝ ⎠ =

2 1

7 6 =

2

42

(iii) Mean ( )x = x1P

1 + x

2P

2 + x

3P

3

(33)


= 20 20 2

0 1 242 42 42

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= 24

42=4

7

(iv) Variance =

22

0

( )x

P x x x

∑ =

2

4(0) 0

7P

⎛ ⎞ ⎜ ⎟⎝ ⎠

2 2

4 4(1) 1 (2) 2

7 7P P

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 20 16 20 9 2 100

42 49 42 49 42 49

= 50

147

28. f(x) = 4 3 4,

3 4 3

xx R

x

f is one-one :

Let 1 2

4,

3x x R and f(x

1) = f(x

2)

1

1

4 3

3 4

x

x

=

2

2

4 3

3 4

x

x

12x1x

2 + 16x

1 + 9x

2 + 12 = 12x

1x

2 + 9x

1 + 16x

2 + 12

7x1 = 7x

2 x

1 = x

2

f is one-one.

f is onto :

Let 43

k R be any number

f(x) = k 4 3

3 4

xk

x

4x + 3 = 3kx + 4k

x = 4 3

4 3

k

k

Also,4 3

4 3

k

k

=4

3

implies –9 = –16 (which is impossible)

4 3

4 3

kf k

k

⎛ ⎞ ⎜ ⎟⎝ ⎠ i.e., f is onto

The function f is invertible i.e., f–1 exist inverse of f.

Let f–1(x) = k

f(k) = x

4 3

3 4

k

k

= x

f–1(x) = 4 3 4,

4 3 3

xx R

x

f–1(0) = –3

4

(34)


and when

f–1(x) = 2

4 3

4 3

x

x

= 2

4x – 3 = 8 – 6x

10x = 11

x = 11

10

OR

(i) Let (e, f) be the identity element for *

for (a, b) Q × Q, we have

(a, b) * (e, f) = (a, b) = (e, f) * (a, b)

(ae, af + b) = (a, b) = (ea, eb + f)

ae = a, af + b = b, a = ea b = eb + f

e = 1, af = 0, e = 1, b = (I)b + f (∵ a need not be '0')

e = 1, f = 0, e = 1, f = 0

(e, f) = (1, 0) Q × Q

(1, 0) is the identity element of A

(ii) Let (a, b) Q × Q

Let (c, d) Q × Q

such that

(a, b) * (c, d) = (1, 0) = (c, d) * (a, b)

(ac, ad + b) = (1, 0) = (ca, cb + d)

ac = 1, ad + b = 0, ca = 1, cb + d = 0

c = 1

a, d =

ba

, 1

a

⎛ ⎞⎜ ⎟⎝ ⎠ b + d = 0 (a 0)

(c, d) = 1,

b

a a

⎛ ⎞⎜ ⎟⎝ ⎠ (a 0)

for a 0 (a, b)–1 = 1,

b

a a

⎛ ⎞⎜ ⎟⎝ ⎠

29. Let A = (–1, 2)

B = (1, 5)

C = (3, 4)

B(1, 5)

C(3, 4)A( 1, 2)

L M N

We have to find the area of ABC.

(35)


Find equation of line AB y – 5 = 2 5

( 1)1 1

x⎛ ⎞ ⎜ ⎟⎝ ⎠

y – 5 = 31

2x

2y – 10 = 3x – 3

3x – 2y + 7 = 0 ..... (1)

y = 3 7

2

x

Equation of BC y – 4 = 5 43

1 3x

⎛ ⎞ ⎜ ⎟⎝ ⎠

y – 4 = 13

2x

x + 2y – 11 = 0 ..... (2)

y = 11

2

x

Equation of AC :

y – 4 = 2 43

1 3x

⎛ ⎞ ⎜ ⎟⎝ ⎠

y – 4 = 13 2 8 3

2x y x ⇒

x – 2y + 5 = 0 ..... (3)

y = 5

2

x

So, required area =

1 3 3

1 1 1

3 7 11 5

2 2 2

x x xdx dx dx

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠∫ ∫ ∫

=

1 3 32 2 2

1 1 1

1 3 1 17 11 5

2 2 2 2 2 2

x x xx x x

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

= 1 3 3 1 9 1 1 9 1

7 7 33 11 15 52 2 2 2 2 2 2 2 2

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦ ⎣ ⎦

= 114 22 4 24

2 = 1

36 28 42

square unit

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Physics, Chemistry Mathematics...2 = 23 – 12 = 11 (b) As 23 12 Mgcontains even number of protons (12) against 23 11 Nawhich has odd number protons (11), therefore 23 12 Mghas greater

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